Fractional-Linear Inequalities

Here, we will consider inequalities of the form `(ax+b)/(cx+d)>0` or `(ax+b)/(cx+d)<0`, where `a!=0,c!=0`.

Example. Solve inequality `(3x+7)/(2x-7)>5`.

In cases where right-hand side doesn't equal 0, we need to transform inequality into equivalent one, where right-hand side equals 0:

`(3x+7)/(2x-7)>5`;

`(3x+7)/(2x-7)-5>0`;

`(3x+7)/(2x-7)-(5(2x-7))/(2x-7)>0`;

`(3x+7-5(2x-7))/(2x-7)>0`;

`(3x+7-10x+35)/(2x-7)>0`;

`(-7x+42)/(2x-7)>0`; divide both sides by -7:

`(x-6)/(2x-7)<0`;

Fraction is negative in two cases:

  1. Numerator is negative, denominator is positive: `{(x-6<0),(2x-7>0):}`.
  2. Numerator is positive, denominator is negative: `{(x-6>0),(2x-7<0):}`.

Therefore, we have set of two systems (recall that set contains solutions that satisfy at least one system).

Set of solutions of the first system is x<6 and x>3.5, i.e. 3.5<x<6.

Set of solutions of the second system is x>6 and x<3.5, i.e. there are no solutions.

Therefore, set of solutions of initial inequality is interval (3.5,6).